Wow this is some really great info! Thanks Jim and to all others who took
the time to respond!

On a side note - I was trying to join the "Glowbugs" mailing list that I've
seen mentioned on a number of web sites, but I havent gotten any responses
to my to "subscribe" emails. Is this a dead list? If so, are there any
similar current alternatives?

On a second side note - does anyone happen to know what the formula would be
to calculate the capacitance between two metal plates of a given area and
given spacing, having an air dielectric?

thanks again to all!



Dave



"jim rozen" wrote in message
...
In article , David Forsyth says...

Anyone
have any further ideas or insights?

Steel is cheap. This is one reason that
one of the most famous two-tube (+ 1 audio)
regenerative receivers - the national SW3 - was
manufactured with a steel chassis, and a
steel enclosure.

Aluminum is easier to work with obviously.

As a practical matter, I would try to keep
your coils at least one coil diameter away
from the metal chassis, steel or aluminum.

In the SW3 they did this by putting the coils
right in the center of the shield compartmets,
and by standing the coil sockets up on standoffs.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

In article , "David Forsyth"
writes:

On a second side note - does anyone happen to know what the formula would be
to calculate the capacitance between two metal plates of a given area and
given spacing, having an air dielectric?

From the good ol' "Green Bible" (ITT Reference Data for Radio
Engineers, 4th Edition, 1956, page 133), for only two plates,
dimensions in inches, air dielectric:

Capacitance (pFd) = 0.225 x Square Area / spacing

Assuming the plates are the same size and shape.

Len Anderson
retired (from regular hours) electronic engineer person

Thanks Len!



Dave



"Avery Fineman" wrote in message
...
In article , "David Forsyth"
writes:

On a second side note - does anyone happen to know what the formula would
be
to calculate the capacitance between two metal plates of a given area and
given spacing, having an air dielectric?

From the good ol' "Green Bible" (ITT Reference Data for Radio
Engineers, 4th Edition, 1956, page 133), for only two plates,
dimensions in inches, air dielectric:

Capacitance (pFd) = 0.225 x Square Area / spacing

Assuming the plates are the same size and shape.

Len Anderson
retired (from regular hours) electronic engineer person

Thanks Len!



Dave



"Avery Fineman" wrote in message
...
In article , "David Forsyth"
writes:

On a second side note - does anyone happen to know what the formula would
be
to calculate the capacitance between two metal plates of a given area and
given spacing, having an air dielectric?

From the good ol' "Green Bible" (ITT Reference Data for Radio
Engineers, 4th Edition, 1956, page 133), for only two plates,
dimensions in inches, air dielectric:

Capacitance (pFd) = 0.225 x Square Area / spacing

Assuming the plates are the same size and shape.

Len Anderson
retired (from regular hours) electronic engineer person

In article , "David Forsyth"
writes:

On a second side note - does anyone happen to know what the formula would be
to calculate the capacitance between two metal plates of a given area and
given spacing, having an air dielectric?

From the good ol' "Green Bible" (ITT Reference Data for Radio
Engineers, 4th Edition, 1956, page 133), for only two plates,
dimensions in inches, air dielectric:

Capacitance (pFd) = 0.225 x Square Area / spacing

Assuming the plates are the same size and shape.

Len Anderson
retired (from regular hours) electronic engineer person